Free Fall Formulas for Distance Fallen and Speed

Calculates the distance ½gt² fallen and the speed gt reached after being released from rest. Time is in seconds, distance in metres, and speed in m/s. Air resistance is ignored.

Explanation

An object released from rest and pulled down by gravity alone is in free fall. From the time it has been falling you get the distance it has covered and the speed it has reached.

y = \dfrac{1}{2} g t^2

v = g t

$y$ is the distance fallen, in metres (m)
$v$ is the speed after time $t$ , in m/s
$g$ is the acceleration of gravity, about 9.8 m/s² at the Earth's surface
$t$ is the time since release, in seconds (s)

The speed grows in proportion to the time, but the distance grows with the square of it, so each second of a fall covers more ground than the second before.

Example

The defaults are a time of 3 s and gravity of 9.8 m/s².

y = \dfrac{1}{2} \times 9.8 \times 3^2 = 44.1\,\mathrm{m}

v = 9.8 \times 3 = 29.4\,\mathrm{m/s}

In 3 seconds the object falls 44.1 m and reaches 29.4 m/s, which is about 106 km/h.

Notes

Air resistance is ignored. In reality the drag grows as the object speeds up, until it balances gravity and the fall settles at a constant terminal velocity. For a feather or a plastic bag, these formulas are useless.

The object starts from rest. Throw it downwards instead and you need the constant-acceleration formulas, with an initial speed.

Gravity does not care about mass. With no air in the way, a cannonball and a feather fall side by side.

The time must be zero or more, and gravity must be greater than zero.